Examining the Rooms - 第一类斯特灵数

Posted 2020-10-02

---恢复内容开始---

2017-08-10 20:32:37

writer：pprp

 

题意如下：

Recently in Teddy‘s hometown there is a competition named "Cow Year Blow Cow".N competitors had took part in this competition.The competition was so intense that the rank was changing and changing. 
Now the question is: 
How many different ways that n competitors can rank in a competition, allowing for the possibility of ties. 
as the answer will be very large,you can just output the answer MOD 20090126. 
Here are the ways when N = 2: 
P1 < P2 
P2 < P1 
P1 = P2 

InputThe first line will contain a T,then T cases followed. 
each case only contain one integer N (N <= 100),indicating the number of people.OutputOne integer pey line represent the answer MOD 20090126.Sample Input

2
2
3

Sample Output

3
13

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代码如下：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

#define ll long long

using namespace std;

const int N = 110;
const int MOD = 20090126;
ll dp[N][N], ans[N], fac[N];

void init()
{
    fac[1] = 1;
    for(int i = 2; i < N; i++)
        fac[i] = (fac[i-1]*i)%MOD;   //阶乘初始化
    memset(dp, 0, sizeof(dp));
    for(int n = 1; n < N; n++)
    {
        dp[n][1] = 1;
        dp[n][n] = 1;
        for(int k = 2; k < n; k++)
        {
            dp[n][k] = dp[n-1][k-1]+k*dp[n-1][k];
            dp[n][k] %= MOD;
        }
    }
}

int main()
{
    int T, n;
    init();
    cin>>T;
    while(T--)
    {
        cin>>n;
        ll ans = 0;
        for(int i = 1; i <= n; i++)
            ans = (ans + fac[i]*dp[n][i]) % MOD;
        cout<<ans<<endl;
    }

    return 0;
}